88324 control system - solution assignment 7 - 2010

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  • 8/9/2019 88324 Control System - Solution Assignment 7 - 2010

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    Assignment 7

    P6.4

    (a) The closed loop characteristic equation is

    3 2

    ( 40)1 ( ) 1 0

    ( 10)( 20)

    30 200 40 0

    K sGH s

    s s s

    or

    s s s Ks K

    ++ = + =

    + +

    + + + + =

    The Routh array is

    3

    2

    1

    0

    1 200

    30 40

    200 03

    40

    K s K s

    K s

    s K

    +

    Therefore, for stability we require 200 - K/3 > 0 and 40K >0. So, the range of K for stability is0

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    The Routh array is3

    2

    1

    0

    1 392

    24 1099266 0

    10992

    s

    s s

    s

    There are two sign changes in the first column indicating two roots to right of s = -2.Combining the results, we determine that there are two roots located between s = -1and s = -2. The roots of the characteristic equation ares1 = -27.6250 and s 2,3 = -1.1875 20.8082j

    We see that indeed the two roots s 2,3 = -1.1875 20.8082j lie between -1 and -2

    P6.7(a) The closed loop characteristic equation is

    3 2101 (100 10 ) 100 0a a s s KK s KK + + + + =

    The Routh array is3

    2

    1

    0

    1 100 10

    101 100

    100

    a

    a

    a

    KK s KK s

    b s KK s

    +

    Where910

    100 0.101 a

    b KK = + >

    Thus examining the first column, we determine that KK a >0 stabilizes the system

    (b) The tracking error is

    20

    100 100( ) lim (1 ( ))

    sa

    e s s T s s KK

    = =

    We require E(s)100/0.01745 = 5729

    When KKa = 5729, the roots of the characteristic polynomial ares1= -10.15 and s 2,3 = - 45.43 j233.25

  • 8/9/2019 88324 Control System - Solution Assignment 7 - 2010

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  • 8/9/2019 88324 Control System - Solution Assignment 7 - 2010

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    The Routh array is

    4

    3

    2

    1

    0

    1 20 10

    7 24 0

    11610

    7

    10

    K s

    K s

    K K s

    s b s K

    +

    Where116

    ( )(24 ) 707 .

    116( )

    7

    K K K

    b K

    + =

    Setting b>0 yields2784 - 398K - K 2>0

    which holds when- 404.88 < K < 6.876

    Examining the first column, we also find that K0 for stability.Combining all the stability regions, we determine that for stability0 0 or 0 < K

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  • 8/9/2019 88324 Control System - Solution Assignment 7 - 2010

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    AP6.3)(a) The steady-state tracking error to a step input is

    0lim (1 ( )) ( ) 1 (0) 1 ss se s T s R s T

    = = =

    We want

    1 0.05 <

    This yields the bounds for 0.95 < < 1.05.the Routh array is

    3

    2

    1

    0

    1

    1 1

    0

    1

    s

    sb s

    s

    +

    where

    2 11

    b

    + =

    +

    Therefore, using the condition that b>0, we obtain the stability range for :

    >0.618

    (c) Choosing = 1 satisfies both the steady-state tracking requirement and stabilityrequirement.